Eigenfunction Matching for a Semi-Infinite Floating Elastic Plate

Introduction

We show here a solution for a semi-infinite Floating Elastic Plate on Finite Depth. The problem was solved by Fox and Squire 1994 but the solution method here is slightly different.

Equations

We consider the problem of small-amplitude waves which are incident on a semi-infinite floating elastic plate occupying water surface for [math]\displaystyle{ x\gt 0 }[/math]. The submergence of the plate is considered negligible. We assume that the problem is invariant in the [math]\displaystyle{ y }[/math] direction, although we allow the waves to be incident from an angle. We also assume that the plate edges are free to move at each boundary, although other boundary conditions could easily be considered using the methods of solution presented here. We begin with the Frequency Domain Problem for a semi-infinite Floating Elastic Plates in the non-dimensional form of Tayler 1986 (Dispersion Relation for a Floating Elastic Plate)

[math]\displaystyle{ \begin{matrix} \left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial z^2} - k_y^2\right) \phi = 0, \;\;\;\; \mbox{ for } -h \lt z \leq 0, \end{matrix} }[/math] [math]\displaystyle{ \begin{matrix} \frac{\partial \phi}{\partial z} = 0, \;\;\;\; \mbox{ at } z = - h, \end{matrix} }[/math] [math]\displaystyle{ \begin{matrix} \left( \beta \left(\frac{\partial^2}{\partial x^2} - k^2_y\right)^2 - \gamma\alpha + 1\right)\frac{\partial \phi}{\partial z} - \alpha\phi = 0, \;\;\;\; \mbox{ at } z = 0, \;\;\; x \geq 0, \end{matrix} }[/math] [math]\displaystyle{ \begin{matrix} \frac{\partial \phi}{\partial z} - \alpha\phi = 0, \;\;\;\; \mbox{ at } z = 0, \;\;\; x \leq 0, \end{matrix} }[/math]

where [math]\displaystyle{ \alpha = \omega^2 }[/math], [math]\displaystyle{ \beta }[/math] and [math]\displaystyle{ \gamma }[/math] and the stiffness and mass constant for the plate. The conditions at the edge of the plate os

[math]\displaystyle{ \begin{matrix} \left(\frac{\partial^3}{\partial x^3} - (2 - \nu)k^2_y\frac{\partial}{\partial x}\right) \frac{\partial\phi}{\partial z}= 0, \;\;\;\; \mbox{ at } z = 0, \;\;\; \mbox{ for } x = 0, \end{matrix} }[/math] [math]\displaystyle{ \begin{matrix} \left(\frac{\partial^2}{\partial x^2} - \nu k^2_y\right)\frac{\partial\phi}{\partial z} = 0, \;\;\;\;\mbox{ at } z = 0, \;\;\; \mbox{ for } x = 0. \end{matrix} }[/math]

Method of solution

Eigenfunction expansion

We will solve the system of equations using an Eigenfunction Matching Method. The method was developed by Fox and Squire 1994. The separation of variables for the left hand region where there is open water is described in Eigenfunction Matching Method for a Semi-Infinite Dock and we consider here only the separation of variables in the plate covered region.

Separation of variables under the Plate

The potential velocity can be written in terms of an infinite series of separated eigenfunctions under each elastic plate, of the form [math]\displaystyle{ \phi = e^{\kappa_\mu x} \cos(k_\mu(z+h)). }[/math] If we apply the boundary conditions given we obtain the Dispersion Relation for a Floating Elastic Plate

[math]\displaystyle{ \begin{matrix} k_\mu\tan{(k_\mu h)}= & -\frac{\alpha}{\beta_\mu k_\mu^{4} + 1 - \alpha\gamma_\mu} \end{matrix} }[/math]

Solving for [math]\displaystyle{ k_\mu }[/math] gives a pure imaginary root with positive imaginary part, two complex roots (two complex conjugate paired roots with positive imaginary part in all physical situations), an infinite number of positive real roots which approach [math]\displaystyle{ {n\pi}/{h} }[/math] as [math]\displaystyle{ n }[/math] approaches infinity, and also the negative of all these roots (Dispersion Relation for a Floating Elastic Plate) . We denote the two complex roots with positive imaginary part by [math]\displaystyle{ k_\mu(-2) }[/math] and [math]\displaystyle{ k_\mu(-1) }[/math], the purely imaginary root with positive imaginary part by [math]\displaystyle{ k_\mu(0) }[/math] and the real roots with positive imaginary part by [math]\displaystyle{ k_\mu(n) }[/math] for [math]\displaystyle{ n }[/math] a positive integer. The imaginary root with positive imaginary part corresponds to a reflected travelling mode propagating along the [math]\displaystyle{ x }[/math] axis. The complex roots with positive imaginary parts correspond to damped reflected travelling modes and the real roots correspond to reflected evanescent modes. In a similar manner, the negative of these correspond to the transmitted travelling, damped and evanescent modes respectively. The coefficient [math]\displaystyle{ \kappa_\mu }[/math] is

[math]\displaystyle{ \kappa_\mu(n) = \sqrt{k_\mu(n)^2 + k_y^2}, }[/math]

where the root with positive real part is chosen or if the real part is negative with negative imaginary part.

Expressions for the potential velocity

We now expand the potential in the two regions using the separation of variables solution. We always include the two complex and one imaginary root under the plate and truncate the expansion at [math]\displaystyle{ M }[/math] real roots on both sides. The potential [math]\displaystyle{ \phi }[/math] can now be expressed as the following sum of eigenfunctions:

[math]\displaystyle{ \phi \approx \left\{ \begin{matrix} { Ie^{-\kappa_{0}x}\frac{\cos(k_0(z+h))}{\cos(k_0h)} }+ { \qquad \qquad \sum_{n=0}^{M}R_n e^{\kappa_n x} \frac{\cos(k_n(z+h))}{\cos(k_n h)} },& \mbox{ for } x \lt 0,\\ { \sum_{n=-2}^{M}T_n e^{-\kappa^\mu_n x} \frac{\cos(k^\mu_n(z+h))}{\cos(k^\mu_n h)} } \mbox{ for } x \gt 0, \end{matrix} \right. }[/math]

Note that the coefficients are normalised by the potential at the free surface rather than at the bottom surface.

Solving via eigenfunction matching

To solve for the coefficients, we require as many equations as we have unknowns. We derive the equations from the free edge conditions and from imposing conditions of continuity of the potential and its derivative in the [math]\displaystyle{ x }[/math]-direction at the plate-water boundary. We impose the latter condition by taking inner products with respect to the orthogonal functions [math]\displaystyle{ \cos \frac{m\pi}{h}(z+h) }[/math], where [math]\displaystyle{ m }[/math] is a natural number. The main reason we use these functions is because this is the method used in the solution for multiple plates. However, if would be better in this simple problem to use the eigenfunctions for the water.

Taking inner products leads to the following equations

[math]\displaystyle{ \begin{matrix} { \int_{-h}^0 \phi(0^-,z)\cos \frac{m\pi}{h}(z+h) \, dz } &=& { \int_{-h}^0 \phi(0^+,z)\cos \frac{m\pi}{h}(z+h) \, dz }\\ { \int_{-h}^0 \frac{\partial\phi}{\partial x}(0^-,z) \cos \frac{m\pi}{h}(z+h) \, dz } &=& { \int_{-h}^0 \frac{\partial\phi}{\partial x}(0^+,z) \cos \frac{m\pi}{h}(z+h) \, dz } \end{matrix} }[/math]

where [math]\displaystyle{ m\in[0,M] }[/math] and [math]\displaystyle{ \phi_\mu }[/math] denotes the potential under the [math]\displaystyle{ \mu }[/math]th plate, i.e. the expression for [math]\displaystyle{ \phi }[/math] valid for [math]\displaystyle{ l_\mu \lt x\lt r_\mu }[/math]. The remaining equations to be solved are given by the two edge conditions satisfied at both edges of each plate

[math]\displaystyle{ \begin{matrix} { \left(\frac{\partial^3}{\partial x^3} - (2 - \nu)k_y^2\frac{\partial}{\partial x}\right)\frac{\partial\phi_\mu}{\partial z} } &=&0, & \mbox{ for } z = 0 \mbox{ and } x = l_\mu,r_\mu,\\ { \left(\frac{\partial^2}{\partial x^2} - \nu k_y^2\right)\frac{\partial\phi_\mu}{\partial z} } &=&0, & \mbox{ for } z = 0 \mbox{ and } x = l_\mu,r_\mu. \end{matrix} }[/math]

We will show the explicit form of the linear system of equations which arise when we solve these equations. Let [math]\displaystyle{ {\mathbf T} }[/math] be a column vector given by [math]\displaystyle{ \left[T_{-2}, . . ., T_M\right]^{{\mathbf T}} }[/math] and [math]\displaystyle{ {\mathbf R} }[/math] be a column vector given by [math]\displaystyle{ \left[R_0 . . . R_M)\right]^{{\mathbf T}} }[/math].

The equations which arise from matching at the boundary between the water and plate are

[math]\displaystyle{ \begin{matrix} I{\mathbf C} + {\mathbf M}^{+} {\mathbf R} ={\mathbf M}^{-} {\mathbf T}\\ -\kappa_1(0)I\mathbf{C} + {\mathbf N}^{+} {\mathbf R} = {\mathbf N}^{-} {\mathbf T}. \end{matrix} }[/math]

where \begin{matrix} { {\mathbf M}^{+}(m,n) = \int_{-h}^0 \frac{\cos \left(k_n (z+h)\right)}{\cos \left(k_n h\right)} \cos \left(\frac{m\pi}{h}(z+h)\right) \, dz}, \\ { {\mathbf M}^{+}_{R_\mu}(m,n) = \int_{-h}^0 \frac{\cos \left(k_{\mu}(n) (z+h)\right)}{\cos \left(k_{\mu}(n) h\right)} \cos \left(\frac{m\pi}{h}(z+h)\right) \, dz },\\ { {\mathbf M}^{-}_{T_\mu}(m,n) = {\mathbf M}^{+}_{R_\mu}(m,n) }\\ { {\mathbf M}^{-}_{R_\mu}(m,n) = {\mathbf M}^{+}_{T_\mu}(m,n). } \end{matrix}

</math>

[math]\displaystyle{ {\mathbf N}^{+}_{T_\mu} }[/math], [math]\displaystyle{ {\mathbf N}^{+}_{R_\mu} }[/math], [math]\displaystyle{ {\mathbf N}^{-}_{T_\mu} }[/math], and [math]\displaystyle{ {\mathbf N}^{-}_{R_\mu} }[/math] are given by

[math]\displaystyle{ \begin{matrix} {\mathbf N}^{\pm}_{T_\mu}(m,n)= -\kappa_\mu(n){\mathbf M}^{\pm}_{T_\mu}(m,n),\\ {\mathbf N}^{\pm}_{R_\mu}(m,n)= \kappa_\mu(n){\mathbf M}^{\pm}_{R_\mu}(m,n). \end{matrix} }[/math]

[math]\displaystyle{ \mathbf{C} }[/math] is a [math]\displaystyle{ (M+1) }[/math] vector which is given by

[math]\displaystyle{ {\mathbf C}(m)=\int_{-h}^0 \frac{\cos (k_1(0)(z+h))}{\cos (k_1(0)h)} \cos \left(\frac{m\pi}{h}(z+h)\right)\, dz. }[/math]

The integrals in the above equation are each solved analytically. Now, for all but the first and [math]\displaystyle{ \Lambda }[/math]th plate, the edge equation becomes

[math]\displaystyle{ \begin{matrix} {\mathbf E}^{+}_{T_\mu} {\mathbf T}_\mu + {\mathbf E}^{+}_{R_\mu} {\mathbf R}_\mu = 0,\\ {\mathbf E}^{-}_{T_\mu} {\mathbf T}_\mu + {\mathbf E}^{-}_{R_\mu} {\mathbf R}_\mu = 0. \end{matrix} }[/math]

The first and last plates only require two equations, because each has only one plate edge. The equation for the first plate must be modified to include the effect of the incident wave. This gives us

[math]\displaystyle{ \begin{matrix} I \left( \begin{matrix} {\mathbf E}^{+}_{T_1}(1,0)\\ {\mathbf E}^{+}_{T_1}(2,0) \end{matrix} \right) + {\mathbf E}^{+}_{R_1} {\mathbf R}_1 = 0,\\ \end{matrix} }[/math]

and for the [math]\displaystyle{ \Lambda }[/math]th plate we have no reflection so

[math]\displaystyle{ \begin{matrix} {\mathbf E}^{-}_{T_\mu} {\mathbf T}_\mu = 0.\\ \end{matrix} }[/math]

[math]\displaystyle{ {\mathbf E}^{+}_{T_\mu} }[/math], [math]\displaystyle{ {\mathbf E}^{+}_{R_\mu} }[/math], [math]\displaystyle{ {\mathbf E}^{-}_{T_\mu} }[/math] and [math]\displaystyle{ {\mathbf E}^{-}_{R_\mu} }[/math] are 2 by M+3 matrices given by

[math]\displaystyle{ \begin{matrix} {\mathbf E}^{-}_{T_\mu}(1,n) = (\kappa_\mu(n)^2 - (2 - \nu)k_y^2)(k_{\mu}(n)\kappa_\mu(n)\tan{(k_{\mu}(n)h)}),\\ {\mathbf E}^{+}_{T_\mu}(1,n) = (\kappa_\mu(n)^2 - (2 - \nu)k_y^2)(k_{\mu}(n)\kappa_\mu(n)e^{-\kappa_\mu(n)(r_\mu - l_\mu)}\tan{(k_{\mu}(n)h)}),\\ {\mathbf E}^{-}_{R_\mu}(1,n) = (\kappa_\mu(n)^2 - (2 - \nu)k_y^2)(-k_{\mu}(n)\kappa_\mu(n)e^{\kappa_\mu(n)(l_\mu - r_\mu)}\tan{(k_{\mu}(n)h)}),\\ {\mathbf E}^{+}_{R_\mu}(1,n) = (\kappa_\mu(n)^2 - (2 - \nu)k_y^2)(-k_{\mu}(n)\kappa_\mu(n)\tan{(k_{\mu}(n)h)}),\\ {\mathbf E}^{-}_{T_\mu}(2,n) = (\kappa_\mu(n)^2 - \nu k_y^2)(-k_{\mu}(n)\tan{(k_{\mu}(n)h)}),\\ {\mathbf E}^{+}_{T_\mu}(2,n) = (\kappa_\mu(n)^2 - \nu k_y^2)(-k_{\mu}(n)e^{-\kappa_\mu(n)(r_\mu - l_\mu)}\tan{(k_{\mu}(n)h)}),\\ {\mathbf E}^{-}_{R_\mu}(2,n) = (\kappa_\mu(n)^2 - \nu k_y^2)(-k_{\mu}(n)e^{\kappa_\mu(n)(l_\mu - r_\mu)}\tan{(k_{\mu}(n)h)}),\\ {\mathbf E}^{+}_{R_\mu}(2,n) = (\kappa_\mu(n)^2 - \nu k_y^2)(-k_{\mu}(n)\tan{(k_{\mu}(n)h)}).\\ \end{matrix} }[/math]

Now, the matching matrix is a [math]\displaystyle{ (2M+6)\times(\Lambda-1) }[/math] by [math]\displaystyle{ (2M+1)\times(\Lambda -1) }[/math] matrix given by

[math]\displaystyle{ {\mathbf M} = \left( \begin{matrix} {\mathbf M}^{+}_{R_1} & -{\mathbf M}^{-}_{T_2} & -{\mathbf M}^{-}_{R_2} & 0 & 0 & & 0 & 0 & 0 \\ {\mathbf N}^{+}_{R_1} & -{\mathbf N}^{-}_{T_2} & -{\mathbf N}^{-}_{R_2} & 0 & 0 & & 0 & 0 & 0 \\ 0 & {\mathbf M}^{+}_{T_2} & {\mathbf M}^{+}_{R_2} & -{\mathbf M}^{-}_{T_3} & -{\mathbf M}^{-}_{R_3} & & 0 & 0 & 0 \\ 0 & {\mathbf N}^{+}_{T_2} & {\mathbf N}^{+}_{R_2} & -{\mathbf N}^{-}_{T_3} & -{\mathbf N}^{-}_{R_3} & & 0 & 0 & 0 \\ & & \vdots & & & \ddots & \\ 0 & 0 & 0 & 0 & 0 & & {\mathbf M}^{+}_{T_{\Lambda - 1}} & {\mathbf M}^{+}_{R_{\Lambda - 1}} & -{\mathbf M}^{-}_{ T_{\Lambda}} \\ 0 & 0 & 0 & 0 & 0 & & {\mathbf N}^{+}_{T_{\Lambda - 1}} & {\mathbf N}^{+}_{R_{\Lambda - 1}} & -{\mathbf N}^{-}_{T_{\Lambda }} \\ \end{matrix} \right), }[/math]

the edge matrix is a [math]\displaystyle{ (2M+6)\times(\Lambda-1) }[/math] by [math]\displaystyle{ 4(\Lambda-1) }[/math] matrix given by

[math]\displaystyle{ {\mathbf E} = \left( \begin{matrix} {\mathbf E}^{+}_{R_1} & 0 & 0 & 0 & 0 & & 0 & 0 & 0 \\ 0 & {\mathbf E}^{+}_{T_2} & {\mathbf E}^{+}_{R_2} & 0 & 0 & & 0 & 0 & 0 \\ 0 & {\mathbf E}^{-}_{T_2} & {\mathbf E}^{-}_{R_2} & 0 & 0 & & 0 & 0 & 0 \\ 0 & 0 & 0 & {\mathbf E}^{+}_{T_3} & {\mathbf E}^{+}_{R_3} & & 0 & 0 & 0 \\ 0 & 0 & 0 & {\mathbf E}^{-}_{T_3} & {\mathbf E}^{-}_{R_3} & & 0 & 0 & 0 \\ & & \vdots & & & \ddots & \\ 0 & 0 & 0 & 0 & 0 & & {\mathbf E}^{+}_{T_{\Lambda-1}} & {\mathbf E}^{+}_{R_{\Lambda-1}} & 0 \\ 0 & 0 & 0 & 0 & 0 & & {\mathbf E}^{-}_{T_{\Lambda-1}} & {\mathbf E}^{-}_{R_{\Lambda-1}} & 0 \\ 0 & 0 & 0 & 0 & 0 & & 0 & 0 & {\mathbf E}^{-}_{ T_\Lambda} \end{matrix} \right), }[/math]

and finally the complete system to be solved is given by

[math]\displaystyle{ \left( \begin{matrix} {\mathbf M}\\ {\mathbf E}\\ \end{matrix} \right) \times \left( \begin{matrix} {\mathbf R}_1\\ {\mathbf T}_2\\ {\mathbf R}_2\\ {\mathbf T}_3\\ {\mathbf R}_3\\ \vdots\\ {\mathbf T}_{\Lambda-1}\\ {\mathbf R}_{\Lambda-1}\\ {\mathbf T}_{\Lambda} \end{matrix} \right) = \left( \begin{matrix} -I{\mathbf C}\\ \kappa_{1}(0)I{\mathbf C}\\ 0\\ \vdots\\ -IE^{+}_{T_1}(1,0)\\ -IE^{+}_{T_1}(2,0)\\ 0\\ \vdots \end{matrix} \right). }[/math]

The final system of equations has size [math]\displaystyle{ (2M+6)\times (\Lambda - 1) }[/math] by [math]\displaystyle{ (2M+6)\times (\Lambda - 1) }[/math].